Math Is Fun Forum

tony123

Member

Registered: 2007-08-03

Posts: 229

Sum of Powers

Find the last decimal digit of the sum 1^1 + 2^2 + 3^3 + ... + 2001^2001

Identity

Member

Registered: 2007-04-18

Posts: 934

Re: Sum of Powers

Last digit of each

n
1: 1
2: 4
3: 7
4: 6
5: 5
6: 6
7: 3
8: 6
9: 9
10: 0
Sum of last digits = 47

Now simply multiply 47 by 10, then multiply that by 20, then add 1.

Find the last digit of that. Think about why this yields the answer.

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Sum of Powers

You don’t need to find n[sup]n[/sup] for each n. You just need to know that n[sup]5[/sup] has the same last digit as n – so you just work with n (mod 4). The cycle repeats at n = lcm(4,10) = 20. Hence, if you sum 1[sup]1[/sup]+2[sup]2[/sup]+…+20[sup]20[/sup] ten times, you will get back to 0 (mod 10). In other words

Thus the last digit of 1[sup]1[/sup]+2[sup]2[/sup]+…+2001[sup]2001[/sup] is simply the last digit of 2001[sup]2001[/sup], namely 1.

I hope I’m making sense.